Recover from div by 0

div0.c

/* div0.c -- Recover from a division by zero.
** References:
** https://learn.microsoft.com/en-us/cpp/cpp/writing-an-exception-filter
** http://ref.x86asm.net/coder64.html
** https://www.felixcloutier.com/x86/idiv
** AMD64 Architecture Programmer's Manual Volume 3
** Intel Architectures Software Developer's Manual Volume 2
** https://web.archive.org/web/20110304144303/http://www.microsoft.com/msj/0898/bugslayer0898.aspx
*/

#define WIN32_LEAN_AND_MEAN
#include <windows.h>
#include <stdio.h>
#include <stdint.h>

int seh_filter(int code, PEXCEPTION_POINTERS err);

void main(void)
{
	int dividend = 100;
	int zero = 0;
	__try
	{
		int result = dividend / zero; /* <-- Divides by 0. */
		printf("%d / %d = %d\n", dividend, zero, result);
	}
	__except (seh_filter(GetExceptionCode(), GetExceptionInformation()))
	{
		puts("Could not recover from division by 0.");
	}
}

/* Structured exception handler filter. */
int seh_filter(int code, PEXCEPTION_POINTERS err)
{
	if (code == EXCEPTION_INT_DIVIDE_BY_ZERO)
	{
		uint8_t *ip = (uint8_t *)err->ContextRecord->Rip;
		fprintf(stderr, "[division by zero at rip:%p]\n", ip);
		/* Decode the size of a DIV or IDIV to jump over it. */
		if ((*ip & 0xf0) == 0x40) ip += 1; /* Skip REX prefix. */
		if (*ip != 0xf7) return EXCEPTION_EXECUTE_HANDLER; /* Check Opcode. */
		ip += 1;
		if ((*ip & 0x30) != 0x30) return EXCEPTION_EXECUTE_HANDLER; /* Check Reg for 6 or 7. */
		uint8_t mod = *ip >> 6; /* Read ModR/M. */
		uint8_t rm = *ip & 7;
		ip += 1;
		if (mod != 3)
		{
			if (mod == 2) ip += 4; /* Skip disp32. */
			if (mod == 1) ip += 1; /* Skip disp8. */
			if (mod == 0 && rm == 5) ip += 4; /* Skip disp32. */
			if (rm == 4) ip += 1; /* Skip SIB. */
		}
		err->ContextRecord->Rax = err->ContextRecord->Rdx = 0; /* Clear destinations. */
		err->ContextRecord->Rip = (DWORD64)ip; /* Jump. */
		return EXCEPTION_CONTINUE_EXECUTION;
	}
	return EXCEPTION_CONTINUE_SEARCH;
}